Page:Optics.djvu/97

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β=(∆″−r′)²·(1/m∆′−1/∆″)v′.

The whole aberration is therefore

m∆″²/∆′²(∆′−r)²(m/∆−1/∆′)v+(∆″−r′)²·(1/m∆′−1/∆″)v′.

The angles AER, Aer being very nearly the same, we may, without much error, establish that for a particular value of ∆ the aberration varies as v the versed sine of AER, that is, as the square of AR, the radius of the aperture.

Let us examine what kinds of value the aberration in a lens assumes in different cases.

1. For the meniscus or concavo-convex lens we have (r, r′, being both positive.)

The aberration

(A)={m·∆″²/∆′²·(∆′−r)²(m/∆−1/∆′)+(∆″−r′)²(1/m∆′−1/∆″)}v.

Now suppose m=3/2, r=1, r′=5/3, ∆=∞ and therefiore

∆′=	3r=3, ∆″=5.
A=	{−3/2·52/32·22·1/3+102/32(2/3·1/3−1/5)}v
=	−430/81v.

And if v be the versed sine of 2° or .0006, A=−.003, nearly.

Note that the aberration is of a contrary sign to the focal distance, and therefore diminishes it.

2. For the double-concave lens r′ is negative.

A={m·∆″²/∆′²·(∆′−r)²·(m/∆−1/∆′)+(∆″−r′)²·(1/m∆′−1/∆″)}v.